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1 gm of helium having rms velocity of molecules 100 m/s and 4 gm of oxygen having rms velocity of molecules 1000 m/s are mixed in a container which is thermally isolated. What are the rms velocities of helium and oxygen molecules after equilibrium is attained ? |
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Answer» Solution :Given that rms velocity of He molecules is 1000 m/s. If gas is at temperature T, we have `v_("rms")=sqrt((3RT)/(M_(1)))` or `""1000=sqrt((3xx8.314xxT_(1))/(2xx10s^(-3)))` or `""T_(1)=(2xx10^(-3)xx10^(6))/(3xx8.314)=80.186` K It is given that rms velocity of oxygen molecules is also 1000 m/s. If temperature of this gas is `T_(2)`, we have `v_("rms")=sqrt((3RT_(2))/(M_(2)))` or `""1000=sqrt((3xx8.314xxT_(2))/(32xx10^(-3)))` or `""T_(2)=(32xx10^(-3)xx10^(6))/(3xx8.314)=1282.976" K"` It is given that 1 GM of He and 4 gm of `O_(2)` is mixed. If their NUMBER of moles are `n_(1)` and `n_(2)` then `n_(1)=(1)/(2)" and "n_(2)=(4)/(32)=(1)/(8)` We know that gases at different temperature are mixed at constant volume (or in a container), the total internal energy of system remains constant before and after mixing. If in this CASE final temperature of mixture is `T_(f)` then we have `T_(f)=(f_(1)n_(1)T_(1)+f_(2)n_(2)T_(2))/(f_(1)n_(1)+f_(2)n_(2))` Here we have `f_(1)=3,f_(2)=5,n_(1)=(1)/(2),n_(2)=(1)/(8),T_(1)80.186` and `T_(2)=1282.76` Thus final temperature of mixture is given as `T_(f)=(3xx(1)/(2)xx80.186+5xx(1)/(8)xx1282.76)/(3xx(1)/(2)+5xx(1)/(8))` `=(120.28xx801.725)/(2.125)=433.89" K"` Thus final rms velocity of He gas molecules is `v_("rms")=sqrt((3RT_(f))/(M_(1)))` `=sqrt((3xx8.314xx433.89)/(2xx10^(-3)))=2326.17" m"//"s"` Final rms velocity of `O_(2)` gas molecules is `v_("rms")=sqrt((3RT_(f))/(M_(2)))` `=sqrt((3xx8.314xx433.89)/(32xx10^(-3)))=581.54" m"//"s"` |
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